Question 1135704

the quadratic is written in the form {{{y = a(x  -  h)^2 + k}}}, where the vertex is the point ({{{h}}}, {{{k}}})

given:

({{{h}}}, {{{k}}})
= ({{{2}}}, {{{0}}})

and ({{{x}}}, {{{y}}})=({{{4}}}, {{{4}}})

using vertex, you have

{{{y = a(x  -  2)^2 + 0}}}

{{{y = a(x  -  2)^2 }}}


now use given point and calculate {{{a}}}

{{{4= a(4  -  2)^2}}}

{{{4= a(2)^2 }}}

{{{4= 4a }}}

{{{a=1}}}


so your equation is: {{{y = 1(x  -  2)^2 }}}=>{{{y = (x  -  2)^2 }}}



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,0,.12),locate(2,-0.5,V(2,0)),
circle(4,4,.12),locate(4,4,p(4,4)),

 graph( 600, 600, -10, 10, -10, 10, (x  -  2)^2)) }}}